f(x)=[x^2+5]/[x-3] where x not equal to 3
set $\displaystyle f(x) = y$ and make x the subject:
$\displaystyle yx - 3y = x^2 + 5$
$\displaystyle x^2 - yx = -(5+3y)$
$\displaystyle (x+\frac{y}{2})^2 - \frac{y^2}{4} = -5 - 3y$ <-- from completing the square
$\displaystyle (x+\frac{y}{2}) = ±\sqrt{\frac{y^2}{4} - 3y - 5}$
$\displaystyle x = - \frac{y}{2} ±\sqrt{\frac{y^2}{4} - 3y - 5} $
Now set any y to x and set x to $\displaystyle f^{-1}(x)$
$\displaystyle f^{-1}(x) = - \frac{x}{2} ±\sqrt{\frac{x^2}{4} - 3x - 5}$